Hard Further Maths Central limit theorem Questions

Challenging, exam-style Further Maths Central limit theorem questions with worked solutions. Stretch yourself on the hardest central-limit-theorem, sampling-distribution-of-the-mean, distribution-of-a-sum, continuity-correction problems.

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Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The length of a randomly selected steel rod, in millimetres, is modelled by a random variable XX with mean μ=120\mu=120 and standard deviation σ=5\sigma=5. The distribution of XX is not known to be normal. A random sample of n=100n=100 steel rods is taken. Let Xˉ\bar{X} denote the sample mean length. Which of the following statements best explains why Xˉ\bar{X} may be modelled by a normal distribution?
Show worked solution

Worked solution

  1. State the approximating distribution given by the central limit theorem

    Xˉ≈N(120,0.25)\bar{X}\approx N\left(120,0.25\right)

    For large nn the sample mean is approximately normal; the second parameter is the variance.

  2. Write down the population parameters and the sample size

    μ=120,σ2=25,n=100\mu=120,\quad\sigma^{2}=25,\quad n=100

    These are the three quantities the central limit theorem needs.

  3. Find the mean of the sampling distribution

    E(Xˉ)=μ=120E(\bar{X})=\mu=120

    The sample mean is an unbiased estimator of μ\mu.

  4. Find the variance of the sample mean

    Var⁡(Xˉ)=σ2n=25100=0.25\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n}=\frac{25}{100}=0.25

    The variance is divided by nn, not by n\sqrt{n}.

  5. Take the square root to obtain the standard error

    sd⁡(Xˉ)=σn=0.25=0.5000\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}=\sqrt{0.25}=0.5000

    The standard error is σ/n\sigma/\sqrt{n}.

  6. Reject the two standard incorrect standard errors

    σn=0.0500,σ2n=2.5000(both wrong)\frac{\sigma}{n}=0.0500,\qquad\frac{\sigma^{2}}{\sqrt{n}}=2.5000\qquad\text{(both wrong)}

    Neither equals σ/n=0.5000\sigma/\sqrt{n}=0.5000.

  7. Contrast the sample mean with the sample total

    Var⁡(Xˉ)=0.25≠2500=nσ2=Var⁡(∑i=1nXi)\operatorname{Var}(\bar{X})=0.25\neq 2500=n\sigma^{2}=\operatorname{Var}\left(\sum_{i=1}^{n}X_{i}\right)

    The question asks about the mean, so σ2/n\sigma^{2}/n is required.

  8. Recall the statement of the central limit theorem

    Xˉ≈N(μ,σ2n)for large n\bar{X}\approx N\left(\mu,\frac{\sigma^{2}}{n}\right)\quad\text{for large }n

    For a large sample the sample mean is approximately normal whatever the distribution of XX.

  9. Recall the corresponding result for a sample total

    ∑i=1nXi≈N(nμ,nσ2)for large n\sum_{i=1}^{n}X_{i}\approx N\left(n\mu,n\sigma^{2}\right)\quad\text{for large }n

    The total has mean nμn\mu and variance nσ2n\sigma^{2}, not μ\mu and σ2/n\sigma^{2}/n.

  10. Distinguish the variance of the sample mean from its standard deviation

    Var⁡(Xˉ)=σ2n,sd⁡(Xˉ)=σn\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n},\qquad\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}

    The standard error is σ/n\sigma/\sqrt{n}; it is neither σ/n\sigma/n nor σ2/n\sigma^{2}/\sqrt{n}.

  11. Justify the variance of the sample mean

    Var⁡(1n∑i=1nXi)=1n2⋅nσ2=σ2n\operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^{n}X_{i}\right)=\frac{1}{n^{2}}\cdot n\sigma^{2}=\frac{\sigma^{2}}{n}

    Independence makes the variances add, and the factor 1/n1/n is squared.

  12. Justify the mean of the sample mean

    E(Xˉ)=1n∑i=1nE(Xi)=nμn=μE(\bar{X})=\frac{1}{n}\sum_{i=1}^{n}E(X_{i})=\frac{n\mu}{n}=\mu

    The sample mean is an unbiased estimator of the population mean.

  13. Write down the standardising transformation

    Z=Xˉ−μσ/n∼N(0,1)approximatelyZ=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1)\quad\text{approximately}

    Subtract the mean and divide by the standard error.

  14. Recall the symmetry of the standard normal distribution

    Φ(−z)=1−Φ(z)\Phi(-z)=1-\Phi(z)

    Negative zz-values are handled by symmetry about 00.

  15. Recall the complement rule for the upper tail

    P(Z>z)=1−Φ(z)P(Z>z)=1-\Phi(z)

    The upper tail is one minus the cumulative probability.

  16. Select the statement that correctly justifies the approximation

    Xˉ≈N(120,0.25)by the central limit theorem\bar{X}\approx N\left(120,0.25\right)\quad\text{by the central limit theorem}

    The central limit theorem applies because nn is large, whatever the distribution of XX.

Answer
Xˉ≈N(μ,σ2n) by the central limit theorem\bar{X}\approx N\left(\mu,\frac{\sigma^{2}}{n}\right)\text{ by the central limit theorem}
Question 2
9 markschallenging
The mass of a randomly selected ball bearing, in grams, is modelled by a random variable XX with mean μ=52.4\mu=52.4 and standard deviation σ=3.2\sigma=3.2. The distribution of XX is not known to be normal. A random sample of n=125n=125 ball bearings is taken. Let Xˉ\bar{X} denote the sample mean mass. Using the central limit theorem, which of the following is the best estimate of P(52<Xˉ<53)P(52<\bar{X}<53), correct to 44 decimal places? As the approximating normal distribution is continuous, strict and non-strict inequalities give the same value.
Show worked solution

Worked solution

  1. State the approximating distribution given by the central limit theorem

    Xˉ≈N(52.4,0.08192)\bar{X}\approx N\left(52.4,0.08192\right)

    For large nn the sample mean is approximately normal; the second parameter is the variance.

  2. Write down the population parameters and the sample size

    μ=52.4,σ2=10.24,n=125\mu=52.4,\quad\sigma^{2}=10.24,\quad n=125

    These are the three quantities the central limit theorem needs.

  3. Find the mean of the sampling distribution

    E(Xˉ)=μ=52.4E(\bar{X})=\mu=52.4

    The sample mean is an unbiased estimator of μ\mu.

  4. Find the variance of the sample mean

    Var⁡(Xˉ)=σ2n=10.24125=0.08192\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n}=\frac{10.24}{125}=0.08192

    The variance is divided by nn, not by n\sqrt{n}.

  5. Take the square root to obtain the standard error

    sd⁡(Xˉ)=σn=0.08192=0.2862\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}=\sqrt{0.08192}=0.2862

    The standard error is σ/n\sigma/\sqrt{n}.

  6. Reject the two standard incorrect standard errors

    σn=0.0256,σ2n=0.9159(both wrong)\frac{\sigma}{n}=0.0256,\qquad\frac{\sigma^{2}}{\sqrt{n}}=0.9159\qquad\text{(both wrong)}

    Neither equals σ/n=0.2862\sigma/\sqrt{n}=0.2862.

  7. Contrast the sample mean with the sample total

    Var⁡(Xˉ)=0.08192≠1280=nσ2=Var⁡(∑i=1nXi)\operatorname{Var}(\bar{X})=0.08192\neq 1280=n\sigma^{2}=\operatorname{Var}\left(\sum_{i=1}^{n}X_{i}\right)

    The question asks about the mean, so σ2/n\sigma^{2}/n is required.

  8. Standardise the lower endpoint

    z1=52−52.40.2862=−1.3975z_{1}=\frac{52-52.4}{0.2862}=-1.3975

    Subtract the mean and divide by the standard deviation.

  9. Standardise the upper endpoint

    z2=53−52.40.2862=2.0963z_{2}=\frac{53-52.4}{0.2862}=2.0963

    The same standardisation is applied to the other endpoint.

  10. Write the probability as a difference of cumulative probabilities

    P(52<Xˉ<53)=Φ(2.0963)−Φ(−1.3975)P(52<\bar{X}<53)=\Phi(2.0963)-\Phi(-1.3975)

    An interval probability is the difference of two values of Φ\Phi.

  11. Evaluate the upper cumulative probability

    Φ(2.0963)=0.9820\Phi(2.0963)=0.9820

    Read the value from the normal cumulative distribution function.

  12. Evaluate the lower cumulative probability

    Φ(−1.3975)=0.0811\Phi(-1.3975)=0.0811

    Read the value from the normal cumulative distribution function.

  13. Subtract to obtain the probability

    P(52<Xˉ<53)=0.9820−0.0811=0.9008P(52<\bar{X}<53)=0.9820-0.0811=0.9008

    This is the required approximate probability.

  14. Confirm the answer is a valid probability

    0<0.9008<10<0.9008<1

    The value lies strictly between 00 and 11.

  15. Select the option equal to this probability

    P(52<Xˉ<53)≈0.9008P(52<\bar{X}<53)\approx 0.9008

    This is the approximate probability given by the central limit theorem.

Answer
P(52<Xˉ<53)≈0.9008P(52<\bar{X}<53)\approx 0.9008
Question 3
9 markschallenging
The duration of a randomly selected telephone call, in minutes, is modelled by a random variable XX with mean μ=9.6\mu=9.6 and standard deviation σ=1.5\sigma=1.5. The distribution of XX is not known to be normal. A random sample of n=100n=100 telephone calls is taken. Let T=∑i=1nXiT=\sum_{i=1}^{n}X_i denote the total duration of the sample. Using the central limit theorem, which of the following is the best estimate of P(T>970)P(T>970), correct to 44 decimal places? As the approximating normal distribution is continuous, strict and non-strict inequalities give the same value.
Show worked solution

Worked solution

  1. State the approximating distribution given by the central limit theorem

    T≈N(960,225)T\approx N\left(960,225\right)

    For large nn the sample total is approximately normal; the second parameter is the variance.

  2. Write down the population parameters and the sample size

    μ=9.6,σ2=2.25,n=100\mu=9.6,\quad\sigma^{2}=2.25,\quad n=100

    These are the three quantities the central limit theorem needs.

  3. Find the mean of the sample total

    E(T)=nμ=100×9.6=960E(T)=n\mu=100\times 9.6=960

    Expectation is additive, so the total has mean nμn\mu.

  4. Find the variance of the sample total

    Var⁡(T)=nσ2=100×2.25=225\operatorname{Var}(T)=n\sigma^{2}=100\times 2.25=225

    Independent variances add, giving nσ2n\sigma^{2}.

  5. Take the square root to obtain the standard deviation of the total

    sd⁡(T)=nσ2=225=15.0000\operatorname{sd}(T)=\sqrt{n\sigma^{2}}=\sqrt{225}=15.0000

    The standard deviation of the total grows like n\sqrt{n}.

  6. Contrast the sample total with the sample mean

    Var⁡(T)=225≠0.0225=σ2n=Var⁡(Xˉ)\operatorname{Var}(T)=225\neq 0.0225=\frac{\sigma^{2}}{n}=\operatorname{Var}(\bar{X})

    The question asks about the total, so nσ2n\sigma^{2} is required.

  7. Reject the two standard incorrect standard deviations

    nσ=150.0000,σ2n=0.1500(both wrong)n\sigma=150.0000,\qquad\sqrt{\frac{\sigma^{2}}{n}}=0.1500\qquad\text{(both wrong)}

    Neither equals nσ2=15.0000\sqrt{n\sigma^{2}}=15.0000.

  8. Standardise the boundary value

    z=970−96015.0000=0.6667z=\frac{970-960}{15.0000}=0.6667

    Subtract the mean and divide by the standard deviation.

  9. Write the probability in terms of the normal cdf

    P(T>970)=1−Φ(0.6667)P(T>970)=1-\Phi(0.6667)

    An upper-tail probability is one minus the cumulative value.

  10. Evaluate the cumulative probability

    Φ(0.6667)=0.7475\Phi(0.6667)=0.7475

    Read the value from the normal cumulative distribution function.

  11. Subtract from one

    P(T>970)=1−0.7475=0.2525P(T>970)=1-0.7475=0.2525

    This gives the upper-tail probability.

  12. Check the effect of using the distribution of the mean instead of the total

    P(T>970)→0.0000≠0.2525P(T>970)\to 0.0000\neq 0.2525

    The total and the mean have different distributions.

  13. Check the complementary probability

    1−0.2525=0.74751-0.2525=0.7475

    The two probabilities sum to 11, as they must.

  14. Confirm the answer is a valid probability

    0<0.2525<10<0.2525<1

    The value lies strictly between 00 and 11.

  15. Recall the statement of the central limit theorem

    Xˉ≈N(μ,σ2n)for large n\bar{X}\approx N\left(\mu,\frac{\sigma^{2}}{n}\right)\quad\text{for large }n

    For a large sample the sample mean is approximately normal whatever the distribution of XX.

  16. Recall the corresponding result for a sample total

    ∑i=1nXi≈N(nμ,nσ2)for large n\sum_{i=1}^{n}X_{i}\approx N\left(n\mu,n\sigma^{2}\right)\quad\text{for large }n

    The total has mean nμn\mu and variance nσ2n\sigma^{2}, not μ\mu and σ2/n\sigma^{2}/n.

  17. Select the option equal to this probability

    P(T>970)≈0.2525P(T>970)\approx 0.2525

    This is the approximate probability given by the central limit theorem.

Answer
P(T>970)≈0.2525P(T>970)\approx 0.2525
Question 4
9 markschallenging
The mass of a randomly selected screw, in grams, is modelled by a random variable XX with mean μ=25.4\mu=25.4 and standard deviation σ=2\sigma=2. The distribution of XX is not known to be normal. A random sample of n=64n=64 screws is taken. Let Xˉ\bar{X} denote the sample mean mass. Using the central limit theorem, which of the following is the best estimate of P(Xˉ>25.8)P(\bar{X}>25.8), correct to 44 decimal places? As the approximating normal distribution is continuous, strict and non-strict inequalities give the same value.
Show worked solution

Worked solution

  1. State the approximating distribution given by the central limit theorem

    Xˉ≈N(25.4,0.0625)\bar{X}\approx N\left(25.4,0.0625\right)

    For large nn the sample mean is approximately normal; the second parameter is the variance.

  2. Write down the population parameters and the sample size

    μ=25.4,σ2=4,n=64\mu=25.4,\quad\sigma^{2}=4,\quad n=64

    These are the three quantities the central limit theorem needs.

  3. Find the mean of the sampling distribution

    E(Xˉ)=μ=25.4E(\bar{X})=\mu=25.4

    The sample mean is an unbiased estimator of μ\mu.

  4. Find the variance of the sample mean

    Var⁡(Xˉ)=σ2n=464=0.0625\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n}=\frac{4}{64}=0.0625

    The variance is divided by nn, not by n\sqrt{n}.

  5. Take the square root to obtain the standard error

    sd⁡(Xˉ)=σn=0.0625=0.2500\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}=\sqrt{0.0625}=0.2500

    The standard error is σ/n\sigma/\sqrt{n}.

  6. Reject the two standard incorrect standard errors

    σn=0.0313,σ2n=0.5000(both wrong)\frac{\sigma}{n}=0.0313,\qquad\frac{\sigma^{2}}{\sqrt{n}}=0.5000\qquad\text{(both wrong)}

    Neither equals σ/n=0.2500\sigma/\sqrt{n}=0.2500.

  7. Contrast the sample mean with the sample total

    Var⁡(Xˉ)=0.0625≠256=nσ2=Var⁡(∑i=1nXi)\operatorname{Var}(\bar{X})=0.0625\neq 256=n\sigma^{2}=\operatorname{Var}\left(\sum_{i=1}^{n}X_{i}\right)

    The question asks about the mean, so σ2/n\sigma^{2}/n is required.

  8. Standardise the boundary value

    z=25.8−25.40.2500=1.6000z=\frac{25.8-25.4}{0.2500}=1.6000

    Subtract the mean and divide by the standard deviation.

  9. Write the probability in terms of the normal cdf

    P(Xˉ>25.8)=1−Φ(1.6000)P(\bar{X}>25.8)=1-\Phi(1.6000)

    An upper-tail probability is one minus the cumulative value.

  10. Evaluate the cumulative probability

    Φ(1.6000)=0.9452\Phi(1.6000)=0.9452

    Read the value from the normal cumulative distribution function.

  11. Subtract from one

    P(Xˉ>25.8)=1−0.9452=0.0548P(\bar{X}>25.8)=1-0.9452=0.0548

    This gives the upper-tail probability.

  12. Check the effect of using the incorrect standard error σ/n\sigma/n

    P(Xˉ>25.8)→0.0000≠0.0548P(\bar{X}>25.8)\to 0.0000\neq 0.0548

    Using σ/n\sigma/n gives a different, incorrect answer.

  13. Check the complementary probability

    1−0.0548=0.94521-0.0548=0.9452

    The two probabilities sum to 11, as they must.

  14. Confirm the answer is a valid probability

    0<0.0548<10<0.0548<1

    The value lies strictly between 00 and 11.

  15. Recall the statement of the central limit theorem

    Xˉ≈N(μ,σ2n)for large n\bar{X}\approx N\left(\mu,\frac{\sigma^{2}}{n}\right)\quad\text{for large }n

    For a large sample the sample mean is approximately normal whatever the distribution of XX.

  16. Select the option equal to this probability

    P(Xˉ>25.8)≈0.0548P(\bar{X}>25.8)\approx 0.0548

    This is the approximate probability given by the central limit theorem.

Answer
P(Xˉ>25.8)≈0.0548P(\bar{X}>25.8)\approx 0.0548
Question 5
9 markschallenging
The amount spent by a randomly selected shopper, in pounds, is modelled by a random variable XX with mean μ=32\mu=32 and standard deviation σ=9\sigma=9. The distribution of XX is not known to be normal. A random sample of n=200n=200 shoppers is taken. Let Xˉ\bar{X} denote the sample mean amount spent. Using the central limit theorem, estimate P(31<Xˉ<33)P(31<\bar{X}<33), giving your answer to 44 decimal places. As the approximating normal distribution is continuous, strict and non-strict inequalities give the same value.
Show worked solution

Worked solution

  1. State the approximating distribution given by the central limit theorem

    Xˉ≈N(32,0.405)\bar{X}\approx N\left(32,0.405\right)

    For large nn the sample mean is approximately normal; the second parameter is the variance.

  2. Write down the population parameters and the sample size

    μ=32,σ2=81,n=200\mu=32,\quad\sigma^{2}=81,\quad n=200

    These are the three quantities the central limit theorem needs.

  3. Find the mean of the sampling distribution

    E(Xˉ)=μ=32E(\bar{X})=\mu=32

    The sample mean is an unbiased estimator of μ\mu.

  4. Find the variance of the sample mean

    Var⁡(Xˉ)=σ2n=81200=0.405\operatorname{Var}(\bar{X})=\frac{\sigma^{2}}{n}=\frac{81}{200}=0.405

    The variance is divided by nn, not by n\sqrt{n}.

  5. Take the square root to obtain the standard error

    sd⁡(Xˉ)=σn=0.405=0.6364\operatorname{sd}(\bar{X})=\frac{\sigma}{\sqrt{n}}=\sqrt{0.405}=0.6364

    The standard error is σ/n\sigma/\sqrt{n}.

  6. Reject the two standard incorrect standard errors

    σn=0.0450,σ2n=5.7276(both wrong)\frac{\sigma}{n}=0.0450,\qquad\frac{\sigma^{2}}{\sqrt{n}}=5.7276\qquad\text{(both wrong)}

    Neither equals σ/n=0.6364\sigma/\sqrt{n}=0.6364.

  7. Contrast the sample mean with the sample total

    Var⁡(Xˉ)=0.405≠16200=nσ2=Var⁡(∑i=1nXi)\operatorname{Var}(\bar{X})=0.405\neq 16200=n\sigma^{2}=\operatorname{Var}\left(\sum_{i=1}^{n}X_{i}\right)

    The question asks about the mean, so σ2/n\sigma^{2}/n is required.

  8. Standardise the lower endpoint

    z1=31−320.6364=−1.5713z_{1}=\frac{31-32}{0.6364}=-1.5713

    Subtract the mean and divide by the standard deviation.

  9. Standardise the upper endpoint

    z2=33−320.6364=1.5713z_{2}=\frac{33-32}{0.6364}=1.5713

    The same standardisation is applied to the other endpoint.

  10. Write the probability as a difference of cumulative probabilities

    P(31<Xˉ<33)=Φ(1.5713)−Φ(−1.5713)P(31<\bar{X}<33)=\Phi(1.5713)-\Phi(-1.5713)

    An interval probability is the difference of two values of Φ\Phi.

  11. Evaluate the upper cumulative probability

    Φ(1.5713)=0.9419\Phi(1.5713)=0.9419

    Read the value from the normal cumulative distribution function.

  12. Evaluate the lower cumulative probability

    Φ(−1.5713)=0.0581\Phi(-1.5713)=0.0581

    Read the value from the normal cumulative distribution function.

  13. Subtract to obtain the probability

    P(31<Xˉ<33)=0.9419−0.0581=0.8839P(31<\bar{X}<33)=0.9419-0.0581=0.8839

    This is the required approximate probability.

  14. Confirm the answer is a valid probability

    0<0.8839<10<0.8839<1

    The value lies strictly between 00 and 11.

  15. State the final probability

    P(31<Xˉ<33)≈0.8839P(31<\bar{X}<33)\approx 0.8839

    This is the approximate probability, correct to 44 decimal places.

Answer
P(31<Xˉ<33)≈0.8839P(31<\bar{X}<33)\approx 0.8839

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